Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle.
y = x^3 − 10x, y = 6x
To sketch the region enclosed by the given curves, we need to first plot the graphs of the equations y = x^3 − 10x and y = 6x on the same coordinate plane.
Let's start by finding the intersection points of the two curves. Set y = x^3 − 10x equal to y = 6x:
x^3 − 10x = 6x
Rearranging the equation, we have:
x^3 - 16x = 0
Factoring out x, we get:
x(x^2 - 16) = 0
Setting each factor equal to zero, we have two possibilities:
1. x = 0
2. x^2 - 16 = 0
Solving the second equation, we find:
x^2 = 16
Taking the square root of both sides, we have:
x = ±4
Therefore, the intersection points are (0,0), (-4,-24), and (4,24).
Now we can plot the two curves on a graph.
First, let's look at the y = x^3 − 10x curve. When x = 0, y = 0, so the curve passes through the origin (0,0). For x = ±4, we have y = ±24. Thus, the curve intersects the y-axis at (0,0) and the x-axis at (-4,-24) and (4,24).
Next, let's look at the y = 6x curve. This is simply a straight line passing through the origin (0,0) with a slope of 6.
Now that we have the graphs plotted, we can see that the region enclosed by the curves is bounded by the curves themselves. The line y = 6x is above the curve y = x^3 − 10x for x < -4, and below the curve for x > 4. For -4 < x < 4, the curve is above the line.
To decide whether to integrate with respect to x or y, we need to determine which variable represents the independent variable (the one that changes first). Looking at the graphs, we see that the curves are non-horizontal, and the y-values change as the x-values change. Therefore, we will integrate with respect to x.
Finally, let's draw a typical approximating rectangle within the enclosed region. Choose a random value for x, let's say x = 2. Find the corresponding y-value by substituting x = 2 into y = x^3 − 10x:
y = (2)^3 − 10(2) = 8 - 20 = -12
Now, choose an arbitrary width for the rectangle, let's say Δx = 1. The base of the rectangle will be Δx, and the height will be the difference between the y-values at the left and right edges of the rectangle.
For our rectangle, the left edge is at x = 2 and the right edge is at x = 2 + Δx = 3. Substituting these values into y = x^3 − 10x, we find:
y_left = (2)^3 − 10(2) = -12
y_right = (3)^3 − 10(3) = -9
The height of the rectangle is y_right - y_left = (-9) - (-12) = 3.
Therefore, the rectangle has a base length of Δx = 1 and a height of 3.
We can draw the rectangle with the left edge at x = 2 along the curve y = x^3 − 10x. The top of the rectangle will also touch the line y = 6x at the point (2,12), completing our sketch of the typical approximating rectangle.
Note that this is just one example of a typical approximating rectangle, and the specific dimensions may vary depending on the chosen x-value and width Δx.